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A positive integer's known as a special number if the digits of the number produce a purely growing sequence. Alternatively, 5659 isn't a special number. Same goes for the number 5669. Show that there aren't any 5-digit special numbers that are divisible by 11.

To show that this, I focused on the last digits of the numbers and identified a cycle. My proof:

"To prove that there are no five-digit special numbers divisible by 11, we first need to find the first five-digit number divisible by 11. 11*910 = 10010 As seen above, the first five-digit number divisible by 11 is 10010. Now, we need to list all the possible last two digits of all the five-digit numbers divisible by 11 in order to find a repeating cycle of the last two digits: enter image description here

Key: The yellow highlight represents the last two digits of the numbers. Sequences of digits that are circled in pink are numbers where the last two digits are increasing, but the digit before causes the number to not be increasing. Sequences of digits that are circled in purple are numbers where the last two digits cause the number to not be increasing as it repeats. Sequences of digits that are circled in green colour are there to show that the last two digits return to the beginning of the cycle and to show where all the possibilities of the last two digits terminate. Digits that are written in dark green are the third digits of the five-digit numbers divisible by 11. Sequences of digits that are left plain are numbers where the last two digits are not increasing, resulting in the entire number not being increasing.

In the picture above, it is seen that plain sequences of digits (ones without circles) prevent the five-digit number divisible by 11 from being a special number as the last two digits are not increasing. Sequences of digits that are circled in pink prevent the five-digit number divisible by 11 from being a special number as the third digit of the five digit divisible by 11 (written in green) is larger than the following digits. Sequences of digits that are circled in purple prevent the five-digit number divisible by 11 from being a special number as the last two digits repeat. Therefore, it has been proven that all possible last two digits of all the five-digit divisible by 11 prevent the overall number from being a special number. (Read text circled in orange). This, in turn, proves that no five-digit number divisible by 11 is a special number."

However, I don't know if this is convincing enough as it doesn't use the divisibility rule of 11 in the proof. Any help on finding another proof using the divisibility rule of 11 would be extremely appreciated.

Thanks :)

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  • $\begingroup$ Yes, but I don't really know how to build a proof on that. Is it because some numbers have to be larger than numbers following it for the alternating digit sum to be divisible by 11? $\endgroup$ – Cameron Choi Aug 12 '18 at 12:22
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    $\begingroup$ Some things need clarification here. By purely growing do you mean what is more usually called strictly increasing? If so, how is this sequence to be defined? Obviously, not all its terms can always be single digits, but how is it to be generated or interpreted? Is $12345678,$ for example, to be read as the sequence of first eight positive integers or as the sequence $12, 34, 56, 78,$ which is also strictly increasing? $\endgroup$ – Allawonder Aug 12 '18 at 12:30
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    $\begingroup$ Yes, it does have to be strictly increasing (sorry for not clarifying properly). When assessing if they are strictly increasing, they are all meant to be read as one-digit positive integers (they are meant to be read as separate digits). Hence, 12345678 is meant to be read as the sequence of eight positive integers. $\endgroup$ – Cameron Choi Aug 12 '18 at 12:34
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    $\begingroup$ @Allawonder To answer this question, it does not matter whether we have infinite or finite many "special" numbers. $\endgroup$ – Peter Aug 12 '18 at 12:43
  • $\begingroup$ @Peter Perhaps it doesn't, but you'll at least agree that a motivation, or at least context, often makes a problem more amenable to solution. $\endgroup$ – Allawonder Aug 12 '18 at 13:09
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Suppose $$abcde$$ is special and divisble by $11$. Then, the number $$a+c+e-b-d$$ must be divisible by $11$ The number $a+c+e-b-d$ must be positive because of $e>d$ and $c>b$

The maximum sum $a+c+e$ is given by $a=5$ , $c=7$ , $e=9$ , being smaller than $22$.

Hence, we must have $$e=(b-a)+(d-c)+11$$ This is obviously impossible because of $b>a$ and $d>c$

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    $\begingroup$ Since this proof is somewhat cumbersome , you can show the claim as follows somewhat easier : We have $$(a-b)+(c-d)+e<e$$ hence the sum cannot be $11$ or more. But we also have $$(e-d)+(c-b)+a>0$$ hence the sum must be positive. It follows that $$a-b+c-d+e$$ and therefore $abcde$ cannot be divisible by $11$ $\endgroup$ – Peter Aug 12 '18 at 12:40

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